Question:
Light of wavelength \(5000\text{ \AA}\) falls on a sensitive plate with photoelectric work function of \(1.9\text{ eV}\). The kinetic energy of the emitted photoelectron will be
Light of wavelength \(5000\text{ \AA}\) falls on a sensitive plate with photoelectric work function of \(1.9\text{ eV}\). The kinetic energy of the emitted photoelectron will be
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For photoelectric effect, use \(K_{\max}=\frac{1240}{\lambda(\text{nm})}-\phi\), where \(\lambda\) is in nanometer.
Updated On: May 7, 2026
- \(0.58\text{ eV}\)
- \(2.48\text{ eV}\)
- \(1.24\text{ eV}\)
- \(1.16\text{ eV}\)
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The Correct Option is A
Solution and Explanation
The wavelength of incident light is:
\[
\lambda=5000\text{ \AA}.
\]
Convert angstrom into nanometer:
\[
1\text{ \AA}=0.1\text{ nm}.
\]
So:
\[
5000\text{ \AA}=500\text{ nm}.
\]
The energy of a photon in electron volt is:
\[
E=\frac{1240}{\lambda(\text{nm})}.
\]
Substitute:
\[
E=\frac{1240}{500}.
\]
\[
E=2.48\text{ eV}.
\]
Work function is:
\[
\phi=1.9\text{ eV}.
\]
Using Einstein's photoelectric equation:
\[
K_{\max}=E-\phi.
\]
\[
K_{\max}=2.48-1.9.
\]
\[
K_{\max}=0.58\text{ eV}.
\]
Therefore, the kinetic energy of the emitted photoelectron is:
\[
0.58\text{ eV}.
\]
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